Question

solve the equation \\(log_{2}(x^{2}-12x + 43)=3\\). (1 point)\\(\bigcirc\\ x = 6\pm i\sqrt{7}\\)\\(\bigcirc\\ x=-5, -7\\)\\(\bigcirc\\ x = 5, 7\\)\\(\bigcirc\\ x = 6\pm\sqrt{2}\\)

Explanation:

Step1: Convert logarithmic to exponential form

Recall that if $\log_{a}b = c$, then $b = a^{c}$. So for $\log_{2}(x^{2}-12x + 43)=3$, we have $x^{2}-12x + 43=2^{3}$.

Step2: Simplify the right - hand side

Calculate $2^{3}=8$. So the equation becomes $x^{2}-12x + 43 = 8$.

Step3: Rearrange the equation to standard quadratic form

Subtract 8 from both sides: $x^{2}-12x+43 - 8=0$, which simplifies to $x^{2}-12x + 35 = 0$.

Step4: Factor the quadratic equation

We need two numbers that multiply to 35 and add up to - 12. The numbers are - 5 and - 7. So $x^{2}-12x + 35=(x - 5)(x - 7)=0$.

Step5: Solve for x

Set each factor equal to zero:

• If $x - 5=0$, then $x = 5$.
• If $x - 7=0$, then $x = 7$.

Answer:

$x = 5,7$ (corresponding to the option "x = 5,7")